Question

The eccentric angle of point on the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$ at a distance of $\frac{5}{4}$ unit from the focus on the positive $x$-axis is

Answer 1

Given equation of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$

$\Rightarrow a=2,b=\sqrt{3}$

Let $P(2\cos \theta ,\sqrt{3}\sin \theta )$ be a point on ellipse.

Focus is at $(1,0)$

$\Rightarrow {(2\cos \theta -1)}^2+{(\sqrt{3}\sin \theta -0)}^2={\left(\frac{5}{4}\right)}^2$

$\Rightarrow 4{\cos }^2\theta +1-4\cos \theta +3{\sin }^2\theta ={\left(\frac{5}{4}\right)}^2$

$\Rightarrow 3{\cos }^2\theta +{\cos }^2\theta +1-4\cos \theta +3{\sin }^2\theta ={\left(\frac{5}{4}\right)}^2$

$\Rightarrow 3({\cos }^2\theta +{\sin }^2\theta )+{\cos }^2\theta +1-4\cos \theta ={\left(\frac{5}{4}\right)}^2$

$\Rightarrow 3+{\cos }^2\theta +1-4\cos \theta ={\left(\frac{5}{4}\right)}^2$

since ${\cos }^2\theta +{\sin }^2\theta =1$

$\Rightarrow {\cos }^2\theta +4-4\cos \theta ={\left(\frac{5}{4}\right)}^2$ since ${\cos }^2\theta +{\sin }^2\theta =1$

$\Rightarrow {(\cos \theta -2)}^2={\left(\frac{5}{4}\right)}^2$

$\Rightarrow (\cos \theta -2)=\pm \frac{5}{4}$

$\Rightarrow \cos \theta =\pm \frac{5}{4}+2=\frac{13}{4},\frac{3}{4}$

But Range of $\cos \theta$ cannot be greater than $1$.Thus, $\cos \theta \ne \frac{13}{4}$

$\therefore \cos \theta =\frac{3}{4}$

$\therefore \theta ={\cos }^{-1}\frac{3}{4}$